Question: Solve the Initial value problem:

$\displaystyle (5t^4y^{-3}+7t^6y^7)dt + (-3t^5y^{-4} + 7t^7y^6)dy = 0, y(1) = 1 $.

Express your answer in the form $\displaystyle F(t,y) = 2 $, where $\displaystyle F(t,y) $ has no constant term.

$\displaystyle F(t,y)= $ __________________ $\displaystyle = 2 $

Attempt at the question:

First of all I checked the terms to see if they were exact and they are not. I took the integral with respect to $\displaystyle t $ for $\displaystyle M(t,y) $ which is the first part of the equation and it came out to be $\displaystyle t^5y^{-3} + t^7y^7 + k(y) $. Afterwards I took the partial with respect to y of that integral we just got which was: $\displaystyle -3t^5y^{-4} + 7t^6y^7 + k'(y) $. I now would equate both these equations together and would cancel terms. The result gives me $\displaystyle k'(y) = 0 $. So, my equation $\displaystyle F(t,y) = t^5y^{-3} + t^7y^7 $. But my question now is, how do I solve using the initial values since my equation has both terms $\displaystyle t $ and $\displaystyle y $?

Thanks for everyone who contributes!